PROBLEM:
You are given two arrays (without duplicates) nums1
and nums2
where nums1
’s elements are subset of nums2
. Find all the next greater numbers for nums1
's elements in the corresponding places of nums2
.
The Next Greater Number of a number x in nums1
is the first greater number to its right in nums2
. If it does not exist, output -1 for this number.
Example 1:
Input: nums1 = [4,1,2], nums2 = [1,3,4,2]. Output: [-1,3,-1] Explanation: For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1. For number 1 in the first array, the next greater number for it in the second array is 3. For number 2 in the first array, there is no next greater number for it in the second array, so output -1.
Example 2:
Input: nums1 = [2,4], nums2 = [1,2,3,4]. Output: [3,-1] Explanation: For number 2 in the first array, the next greater number for it in the second array is 3. For number 4 in the first array, there is no next greater number for it in the second array, so output -1.
Note:
- All elements in
nums1
andnums2
are unique. - The length of both
nums1
andnums2
would not exceed 1000.
SOLVE:
class Solution:
def nextGreaterElement(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
stack=[]
dict_nums2={}
for num in nums2:
while len(stack) and stack[-1]<num:
dict_nums2[stack[-1]]=num
del stack[-1]
stack.append(num)
res=[]
for num in nums1:
res.append(-1 if num not in dict_nums2 else dict_nums2[num]) #python3已经放弃了has_key方法
return res
思路:用一个堆栈保存nums2中的降序数列,然后按下标顺序遍历剩下的数字n,把比它小的数字pop出来,并且保存它和n的键值对,最后遍历nums1,如果元素在字典中存在,则为键对应值否则为-1.